2008 AIME II Problems/Problem 9
Problem
A particle is located on the coordinate plane at 
. Define a move for the particle as a counterclockwise rotation of 
radians about the origin followed by a translation of 
units in the positive 
-direction. Given that the particle's position after 
moves is 
, find the greatest integer less than or equal to 
.
Solution
Solution 1
Headline text
Let P(x, y) be the position of the particle on the xy-plane, r be the length OP where O is the origin, and "a" be the inclination of OP to the x-axis. If Q(x', y') is the position of the particle after a move from P, then x'=rcos(\pi/4+a)+10 = \sqrt{2}(x - y)/2 + 10 and y' = rsin(\pi/4+a) = \sqrt{2}(x + y)/2. Let (xn, yn) be the position of the particle after the nth move, where x0 = 5 and y0 = 0. Then x(n+1) + y(n+1) = \sqrt{2}(xn)+10, x(n+1) - y(n+1) = -\sqrt{2}(yn)+10. This implies x(n+2) = -yn + 5\sqrt{2}+ 10, y(n+2)=xn + 5\sqrt{2}. Substituting x0 = 5 and y0 = 0, we have x8 = 5 and y8 = 0 again for the first time. p = x150 = x6 = -5\sqrt{2} and q = y150 = y6 = 5 + 5\sqrt{2}. Hence the final answer is
Solution 2
Let the particle's position be represented by a complex number. The transformation takes 
to 
where 
and 
. We let 
and 
so that we want to find 
.
Basically, the thing comes out to
Notice that
Furthermore, 
. Thus, the final answer is
See also
| 2008 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
